Question

A man measures the time period of a simple pendulum inside a stationary lift and finds it to be $10\text{sec}$. If the lift accelerates upwards with an acceleration $\frac{g}{4}$, then find the time period of the pendulum.

• A. $8\sqrt{5}\text{sec}$
• B. $4\sqrt{5}\text{sec}$
• C. $10\sqrt{5}\text{sec}$
• D. $2\sqrt{5}\text{sec}$

Answer 1

The correct option is B $4\sqrt{5}\text{sec}$

In a stationary lift $T=2\pi \sqrt{\frac{l}{g}}=10\text{seconds}...\left(1\right)$
For a lift accelerated upwards, the effective value of acceleration due to gravity:
$g^{\prime }=g+a$
Replacing $g$ by $g^{\prime }$ in Eq.$\left(1\right)$,
$T^{\prime }=2\pi \sqrt{\frac{l}{(g+a)}}...\left(2\right)$
From Eq. $\left(1\right)$ & $\left(2\right)$
$\frac{T^{\prime }}{T}=\sqrt{\frac{g}{g+a}}$
$\frac{T^{\prime }}{T}=\sqrt{\frac{g}{g+\frac{g}{4}}}=\sqrt{\frac{4}{5}}$
$\therefore T^{\prime }=\frac{2T}{\sqrt{5}}=4\sqrt{5}\text{seconds}$